Find the amplitude and period of the function.

[tex]\[ y=\frac{4}{3} \sin(2x) \][/tex]

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Answer :

Certainly! Let's analyze the function given:

[tex]\[ y = \frac{4}{3} \sin(2x) \][/tex]

To find the amplitude and period of this sine function, we'll follow these steps:

1. Identify the amplitude:
The amplitude of a sine function [tex]\( y = A \sin(Bx + C) + D \)[/tex] is given by the coefficient [tex]\( A \)[/tex] in front of the sine function. This coefficient represents the maximum value the sine function can reach from its midline (typically from 0).

Given the function:
[tex]\[ y = \frac{4}{3} \sin(2x) \][/tex]
The amplitude [tex]\( A \)[/tex] is:
[tex]\[ A = \frac{4}{3} \][/tex]

2. Determine the period:
The period of a sine function [tex]\( y = A \sin(Bx + C) + D \)[/tex] is determined by the coefficient [tex]\( B \)[/tex] inside the argument of the sine function (multiplied by [tex]\( x \)[/tex]). The standard period of the sine function [tex]\( \sin(x) \)[/tex] is [tex]\( 2\pi \)[/tex]. For a sine function with an altered argument [tex]\( Bx \)[/tex], the period is given by:
[tex]\[ \text{Period} = \frac{2\pi}{B} \][/tex]

For the given function:
[tex]\[ y = \frac{4}{3} \sin(2x) \][/tex]
Here, [tex]\( B = 2 \)[/tex]. Thus, the period [tex]\( P \)[/tex] is:
[tex]\[ \text{Period} = \frac{2\pi}{2} = \pi \][/tex]

In conclusion, for the function [tex]\( y = \frac{4}{3} \sin(2x) \)[/tex]:
- The amplitude is [tex]\( \frac{4}{3} \)[/tex] (approximately 1.3333).
- The period is [tex]\( \pi \)[/tex] (approximately 3.1416).

Thus, the amplitude and period of the function are [tex]\( 1.3333 \)[/tex] and [tex]\( 3.1416 \)[/tex] respectively.

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